. a) Given Ma/Min Wod Poblems (Additional Review) Solutions + f, fo 5 ( ) + f (i) f 0 no solution ( ) (ii) f is undefined when (not pat of domain) Check endpoints: + () f () () + (5) 6 f (5) (5) 4 (min. value) (ma. value) Thee is an absolute ma. pt. at b) Given g sin + cos, fo g cos sin (i) g 5, and an absolute min. pt. at (, ) cos sin cos cos sin. cos sin 0 (ii) g is defined fo all values of cos 0 o sin 0 in the domain, 6 Check values: g sin + cos sin + cos 6 6 6 sin + cos g g (min. value) (ma. value) Thee is an absolute ma. pt. at (, ) and an absolute min. pt. at 6,.
. Let be the numbe of music playes sold. Let p be the pice pe playe. Given (, p ) ( 8000,50) and (, p ) ( 7900,5) 50 5 then m 8000 7900 00 p 50 Fo the pice function, 00 8000 00 p 5000 + 8000 00 p + 000 p + 0 00 Revenue function: R p + 0 00 0 00 + R + 0 50 (i) R + 0 0 (ii) R is defined fo all values in the 50 domain 0 50 6500 (6500) 6500 + 0 $65 00 Find pice: p They should chage $65 pe playe in ode to maimize evenue.
. a) Given C 80000 +.5 + 0.07, The aveage cost function is given by When 000 items ae poduced, 80000 c (000) +.5 + 0.07(000) 000 $6.50/item 80000 +.5 + 0.07 c. The maginal cost is given by C.5 + 0.4 When 000 items ae poduced, C (000).5 + 0.4(000) $5.50/item 80000 b) Fo minimum aveage cost, find the deivative: c + 0.07 Solve c 0 : 80000 0.07 0.07 80000 4000000 000 A poduction level of 000 items will minimize aveage cost. c) When 000 items ae poduced, 80000 c (000) +.5 + 0.07(000) (000) $9.50/item The minimum aveage cost is $9.50/item when 000 items ae poduced.
4. a) Let be the width of the ectangle and let y be the length. shoeline Ma A y given + y 400 y 400 (sub into A y ) ( 400 ) A y 400 ( 0 00) A 400 4 (i) A 400 4 0 (ii) A is defined fo all values of 400 4 in the domain 00 Check: A (0) 400(0) (0) 0 A (00) 400(00) (00) 0000 (ma.) A (00) 400(00) (00) 0 Find y: y 400 (00) 00 The maimum swimming aea is 0000 m when the dimensions ae 00 m 00 m b) Since the citical numbe fom pat a) is no longe pat of the domain ( 0 50) check new endpoints: A (0) 0 A (50) 400(50) (50) 5000 The new maimum aea is 5000 m when the dimensions ae 50 m 00 m.,
5. Let be the adius and h be the height of the can. Since the cost (C) of the mateials is based on the suface aea (A), we need to find the minimum value of h A + h given V h 000 cm 000 h (sub into A) 000 A + 000 A + ( > 0) 000 A 4 000 (i) A 4 0 (ii) A is undefined when 0 000 4 (but >0) 4 000 000 4 5.4 Check using Second Deivative Test: 4000 A 4 + when 5.4, 4000 A (5.4) 4 + (5.4) > 0 a min eists at 5.4 000 Find h: h (5.4) 0.84 To minimize the cost of the metal, the adius should be 5.4 cm and the height should be 0.8 cm
6. Let be the width of the ectangula potion of the window, and let y be the height of the ectangula potion. To admit the geatest amount of light, ma Aea Aea of ectangle + Aea of semi-cicle A y + (adius ) + + 8 y 8 given Peimete y y 4 (sub into A) y A 4 + 8 + 8 A 8 4 8 0.56 + (i) A 8 4 0 (ii) A is defined fo all values of (4 + ) 8 in the domain 8 (4 + ). m Check: A (0) 8(0) (0) (0) 0 m A (.) 8(.) (.) (.) 4.48 m (ma) A (.56) 8(.56) (.56) (.56).79 m Detemine width (.).4 m To admit the geatest amount of light, the width of the window should be.4 m.
7. Let the length of the ectangle be and the width be y, such that (,y) is a point on the ellipse as shown. ma Aea y 4y y given + 4y 4 4 4y 4 4 y (sub into Aea equation) Aea A( y) 4y 4 4y ( 0 y ) A ( y) 4 4 4y + 4y 4 4y 8y 4( 4 4y ) ( 4 8y ) ( y ) 4 4y 4 4 4y 4 4y 4y 6 (i) ( y ) 4 4y 6 A ( y) 0 y 0 y y (ii) A ( y) is undefined when y Find dimensions: 4 4 y Check: A (0) 4(0) 4 4(0) 0 A 4 4 4 4 (ma) A () 4() 4 4() 0 The lagest ectangle which can be inscibed in the given ellipse has an aea of 4 units when the dimensions of the ectangle is units units.
8. Let t be the time (in hous) since noon, when the boats ae closest togethe. Let d be the distance between them. 5t At :00 noon + t hous, the westbound boat has tavelled 5t km and 0 0t the nothbound boat has tavelled 0t km, so d it is 0 0t km fom the dock. Minimize d such that: d ( 5t ) + ( 0 0t ) 65t + 400 800t + 400t 05 800 + 400 t t ( 0 t ) dd Find the deivative: d 050t 800 dt dd 050t 800 050t 800 5t + 0 0t dt d dd 050t 800 (i) 0 dt 5t + 0 0t 050t 800 0 800 t 0.9 050 (ii) dd dt is defined fo all values of t in the domain Check: d (0) 05(0) 800(0) + 400 0 km d (0.9) 05(0.9) 800(0.9) + 400 5.6 km (min.) d () 05() 800() + 400 5 km The boats ae closest togethe at 0.9 hous afte noon, ie. at appoimately : pm. Note: 0.9 60 minutes
9. Let (00 ) be the distance the cable should be un on land. Let y be the distance the cable should be un unde wate. Fom the diagam, y +0000 y +0000 Minimize cost such that C 40 00 + 80y ( 0 00) 48000 40 + 80 + 0000 Find the deivative: C 40 + 40( + 0000) ( ) 40 + 80 + 0000 00 m P y 00 Q (i) C 40 + 80 0 + 0000 80 40 + 0000 80 40 + 0000 0.5 + 0000 0.5 + 0000 0.5 + 500 0.75 500 0000 00 57.7 (ii) C is defined fo all values in domain Check: C (0) 48000 40(0) + 80 (0) + 0000 $56000 C (57.7) 48000 40(57.7) + 80 (57.7) + 0000 $5498.0 (min) C (00) 48000 40(00) + 80 (00) + 0000 $96.76 To minimize the cost of laying the cable, it should be installed fom Q to a point 4 m east, then 5.5 m unde wate to point P.
0. a) Define points E, F, G, and H as shown in diagam. Note: AEH CGF θ ( 0 θ ) Aea of ABCD AB BC ( AE + EB ) ( BF + FC ) B E F C cm θ 5 cm A H D G Fom AEH, Fom FCG, AE cos 5 θ FC sin 5 θ AE 5cos θ FC 5sin θ Fom EBF, BEF θ EB BF EB sin θ BF cos θ EB sin θ BF cos θ cos ( θ ) sin ( θ ) This step uses Complementay Identities (see pg. 9 of tet) Fom above, Aea of ABCD ( AE + EB ) ( BF + FC ) ( 5cos θ + sin θ)( cos θ + 5sin θ ) 5cos θ + 5sin θcos θ + 9sin θcos θ + 5sin θ 5 sin θ + cos θ + 4sin θcos θ 5() + 7(sin θcos θ ) A( θ ) 5 + 7 sin θ (i) da 4cos 0 d θ θ cos θ 0 θ θ ad 4 da 4cos d θ θ (ii) da is defined fo all values in domain dθ
Check: A(0) 5 + 7 sin[ (0) ] 5 cm A ( 4 ) 5 + 7 sin ( 4 ) A 5 + 7 sin 5 cm cm (ma) The maimum aea is cm when θ adians. 4 b) Fom pat a), in ectangle ABCD, length AE + EB width BF + FC When θ 4 5cos θ + sin θ cos θ + 5sin θ length 5cos + sin width cos + 5sin 4 4 4 4 5 + + 5 8 4 8 4 Fo maimum aea, the dimensions of ectangle ABCD should be 4 cm 4 cm.